## Chemistry (7th Edition)

$[H_3O^+] = [OH^-] = 2.345 \times 10^{-7}M$
Neutral solution: $[H_3O^+] = [OH^-]$ So, we can use a unknown named "x", that has this value: $[H_3O^+] = [OH^-] = x$ Now, we can use the "Kw" equation: $[H_3O^+] \times [OH^-] = K_w$ $x \times x = 5.5 \times 10^{-14}$ $x^2 = 5.5 \times 10^{-14}$ $x = \sqrt {5.5 \times 10^{-14}}$ $x = 2.345 \times 10^{-7}$ So: $[H_3O^+] = [OH^-] = 2.345 \times 10^{-7}M$